How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right or is there a simple trick?
Best Answer
$$n^3+11n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers}}+12n$$
See The product of n consecutive integers is divisible by n factorial
OR The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)